Could anyone help me work through the following set of equations:
I found these online at
http://marian.fsik.cvut.cz/~bodnar/PragueSum_2012/Staquet_1-2.pdf
and I'm trying to work through them. I get the basics from this i.e. that if a water particle moves vertically it will be subjected to a buoyancy force due to it being surrounded by water of different density. However, I don't understand how the solution has worked i.e. how they went from one stage to the next. Could anyone provide more information about how this solution works?
My main aim here is to figure out how the Brunt vaisala buoyancy frequency is derived.
Answer
Let start with Newton's second law in one dimension, then we get
$$\sum F_z = m \frac{\partial^2\zeta}{\partial t^2} $$
There are two forces acting on the particle
- a downward force due to gravity: $m g = \rho(z) V g$
- an upward buoyant force: $m_{displaced} g = \rho(z+\zeta) g$
Fill things in to get: $$\rho(z) \frac{\partial^2\zeta}{\partial t^2} = -(\rho(z)-\rho(z+\zeta) ) g $$
You can approximate the last term with a linearisation using a Taylor expansion
$$\rho(z+\zeta)\approx\rho(z)+\frac{d\rho}{dz}\zeta+\mathcal{O}(\zeta^2)$$
You see that the term containing $\rho(z)$ will cancel out, thus
$$\rho(z) \frac{\partial^2\zeta}{\partial t^2} = \frac{d\rho}{dz}\zeta g $$
Or, two rewrite it in the simple harmonic oscillator language
$$ \frac{\partial^2\zeta}{\partial t^2} - \left(\frac{g}{\rho(z)} \frac{d\rho}{dz}\right)\zeta = 0 $$
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